The integral \int_\pi \over 12^\pi \over 4 \,\,8\cos 2x \over ( \tan x + \cot x )^3 \,dx equals :

Definite Integration PYQ — JEE Main 2017 · 8 Apr · Shift 1 · Q43 with Solution

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Correct answer: A

We need to evaluate $I=\int_{\pi/12}^{\pi/4} \frac{8\cos 2x}{(\tan x+\cot x)^3}\,dx.$

Use

=\frac{\sin^2 x+\cos^2 x}{\sin x\cos x} =\frac{1}{\sin x\cos x}.$$ Hence, $$(\tan x+\cot x)^3=\frac{1}{\sin^3 x\cos^3 x}.$$ So the integrand becomes $$8\cos 2x\cdot \sin^3 x\cos^3 x.$$ Therefore, $$I=\int_{\pi/12}^{\pi/4} 8\cos 2x\,\sin^3 x\cos^3 x\,dx.$$ --- 2. **Use a standard identity** Since $$\sin x\cos x=\frac{1}{2}\sin 2x,$$ we get $$\sin^3 x\cos^3 x=(\sin x\cos x)^3=\left(\frac{1}{2}\sin 2x\right)^3=\frac{1}{8}\sin^3 2x.$$ Thus, $$8\cos 2x\sin^3 x\cos^3 x=8\cos 2x\cdot \frac{1}{8}\sin^3 2x=\sin^3 2x\cos 2x.$$ So $$I=\int_{\pi/12}^{\pi/4} \sin^3 2x\cos 2x\,dx.$$ --- 3. **Substitute** Let $$u=\sin 2x \implies du=2\cos 2x\,dx \implies \cos 2x\,dx=\frac{du}{2}.$$ Then $$I=\frac{1}{2}\int u^3\,du=\frac{u^4}{8}.$$ Now change limits: - When $x=\pi/12$, $$u=\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}.$$ - When $x=\pi/4$, $$u=\sin\left(\frac{\pi}{2}\right)=1.$$ Hence, $$I=\frac{1}{2}\int_{1/2}^{1} u^3\,du =\frac{1}{2}\left[\frac{u^4}{4}\right]_{1/2}^{1} =\frac{1}{8}\left[1-\left(\frac{1}{2}\right)^4\right].$$ Compute: $$\left(\frac{1}{2}\right)^4=\frac{1}{16}.$$ So, $$I=\frac{1}{8}\left(1-\frac{1}{16}\right)=\frac{1}{8}\cdot \frac{15}{16}=\frac{15}{128}.$$ --- 4. **Check options** The value is $$\boxed{\frac{15}{128}}.$$ So the correct option is **A**.

View written solutionFree

Correct answer: A

We need to evaluate $I=\int_{\pi/12}^{\pi/4} \frac{8\cos 2x}{(\tan x+\cot x)^3}\,dx.$

Use

=\frac{\sin^2 x+\cos^2 x}{\sin x\cos x} =\frac{1}{\sin x\cos x}.$$ Hence, $$(\tan x+\cot x)^3=\frac{1}{\sin^3 x\cos^3 x}.$$ So the integrand becomes $$8\cos 2x\cdot \sin^3 x\cos^3 x.$$ Therefore, $$I=\int_{\pi/12}^{\pi/4} 8\cos 2x\,\sin^3 x\cos^3 x\,dx.$$ --- 2. **Use a standard identity** Since $$\sin x\cos x=\frac{1}{2}\sin 2x,$$ we get $$\sin^3 x\cos^3 x=(\sin x\cos x)^3=\left(\frac{1}{2}\sin 2x\right)^3=\frac{1}{8}\sin^3 2x.$$ Thus, $$8\cos 2x\sin^3 x\cos^3 x=8\cos 2x\cdot \frac{1}{8}\sin^3 2x=\sin^3 2x\cos 2x.$$ So $$I=\int_{\pi/12}^{\pi/4} \sin^3 2x\cos 2x\,dx.$$ --- 3. **Substitute** Let $$u=\sin 2x \implies du=2\cos 2x\,dx \implies \cos 2x\,dx=\frac{du}{2}.$$ Then $$I=\frac{1}{2}\int u^3\,du=\frac{u^4}{8}.$$ Now change limits: - When $x=\pi/12$, $$u=\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}.$$ - When $x=\pi/4$, $$u=\sin\left(\frac{\pi}{2}\right)=1.$$ Hence, $$I=\frac{1}{2}\int_{1/2}^{1} u^3\,du =\frac{1}{2}\left[\frac{u^4}{4}\right]_{1/2}^{1} =\frac{1}{8}\left[1-\left(\frac{1}{2}\right)^4\right].$$ Compute: $$\left(\frac{1}{2}\right)^4=\frac{1}{16}.$$ So, $$I=\frac{1}{8}\left(1-\frac{1}{16}\right)=\frac{1}{8}\cdot \frac{15}{16}=\frac{15}{128}.$$ --- 4. **Check options** The value is $$\boxed{\frac{15}{128}}.$$ So the correct option is **A**.